I have looked far and wide for code for fast sorting of n dimensional arrays by the first element, for example if I had the array:
ray = [[1,2,3,7,5][10,11,12,13,14]]
I would want it to come out as ray = [[1,2,3,5,7][10,11,12,14,13]]
There are several ways to do this. One is zipped = zip(ray) zipped.sort() ray = zip(zipped)
but this is extremely slow. Numpy has a much faster way to do it, but it wasn't immediately apparent.
if the above were a numpy array you could simply do the following: indexes = numpy.argsort(ray[0]) for n in xrange(len(ray)) ray[n] = ray[n][indexes]
I did a time test of the two methods below.
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Copy to clipboard1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | rows, cols = 3, 100000
A = np.random.randn(rows, cols)
B = [list(n) for n in A]
start = time()
i = np.argsort(A[0])
for n in xrange(len(A)):
A[n] = A[n][i]
print time() - start
start = time()
zipped = zip(*B)
zipped.sort()
B = zip(*zipped)
print time() - start
|
The following was printed out
0.0469999313354
0.75
So it does the same operation 16x faster.
Odly, changing the rows to 100 reduces the speedup to only 10x (printed values of 0.764999866486 7.8900001049). I was expecting the opposite
